A088305 -id:A088305 - cp's OEIS frontend

A329157 Expansion of Product_{k>=1} (1 - Sum_{j>=1} j * x^(k*j)).

1, -1, -3, -3, -4, 3, 2, 19, 21, 32, 40, 45, 16, 8, -18, -125, -164, -291, -358, -530, -588, -724, -592, -675, -358, -207, 570, 1201, 2208, 3333, 4944, 6490, 8277, 10492, 11800, 13260, 14328, 14722, 12942, 12075, 5640, 603, -10444, -21120, -39360, -55876, -83488

Offset: 0

Ilya Gutkovskiy, Nov 06 2019

sign

Convolution inverse of A329156.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A003056, A032198, A077285, A088305, A104575, A319668, A329156, A385001.

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i>1, b(n, i-1), 0)-
      add(b(n-i*j, min(n-i*j, i-1))*j, j=`if`(i=1, n, 1..n/i)))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..46);  # Alois P. Heinz, Jul 18 2025

nmax = 46; CoefficientList[Series[Product[(1 - Sum[j x^(k j), {j, 1, nmax}]), {k, 1, nmax}], {x, 0, nmax}], x]
nmax = 46; CoefficientList[Series[Product[(1 - x^k/(1 - x^k)^2), {k, 1, nmax}], {x, 0, nmax}], x]

G.f.: Product_{k>=1} (1 - x^k / (1 - x^k)^2).
G.f.: exp(-Sum_{k>=1} ( Sum_{d|k} 1 / (d * (1 - x^(k/d))^(2*d)) ) * x^k).
G.f.: Product_{k>=1} (1 - x^k)^A032198(k).
G.f.: A(x) = Product_{k>=1} 1 / B(x^k), where B(x) = g.f. of A088305.
a(n) = Sum_{k=0..A003056(n)} (-1)^k * A385001(n,k). - Alois P. Heinz, Jul 18 2025

A382992 Number of compositions of n that have at least 1 part equal to 1 and any part 1 at position k can be k different colors.

Original entry on oeis.org

0, 1, 2, 9, 38, 190, 1105, 7465, 57808, 505924, 4940320, 53248819, 627848796, 8037734786, 111017325240, 1645384681388, 26044845197271, 438499277778649, 7824114643729925, 147476551001252541, 2928074880767250057, 61078483577649281698, 1335438738400978500931

Offset: 0

John Tyler Rascoe, Apr 11 2025

			a(3) = 9 counts: (2, 1_a), (2, 1_b), (1_a, 2), (1_a, 1_a, 1_a), (1_a, 1_a, 1_b), (1_a, 1_a, 1_c), (1_a, 1_b, 1_a), (1_a, 1_b, 1_b), (1_a, 1_b, 1_c).

Alois P. Heinz, Table of n, a(n) for n = 0..450

Cf. A000045, A008275, A011782, A088305, A238351, A240736, A382991.

b:= proc(n, i, t) option remember; `if`(n=0, t, add(
      b(n-j, i+1, `if`(j=1, 1, t))*`if`(j=1, i, 1), j=1..n))
    end:
a:= n-> b(n, 1, 0):
seq(a(n), n=0..22);  # Alois P. Heinz, Apr 23 2025

A_x(N) = {my(x='x+O('x^N)); Vec(-x^2/(1-x-x^2) + sum(i=1,N, prod(j=1,i, j*x + x^2/(1-x))))}
A_x(30)

G.f.: -x^2/(1-x-x^2) + Sum_{i>0} Product_{j=1..i} ( j*x + x^2/(1-x) ).

a(n) = A382991(n) - A000045(n-1).

Edited by Alois P. Heinz, Apr 23 2025

A121461 Triangle read by rows: T(n,k) is the number of nondecreasing Dyck paths of semilength n, having last ascent of length k (1 <= k <= n).

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 8, 3, 1, 1, 21, 8, 3, 1, 1, 55, 21, 8, 3, 1, 1, 144, 55, 21, 8, 3, 1, 1, 377, 144, 55, 21, 8, 3, 1, 1, 987, 377, 144, 55, 21, 8, 3, 1, 1, 2584, 987, 377, 144, 55, 21, 8, 3, 1, 1, 6765, 2584, 987, 377, 144, 55, 21, 8, 3, 1, 1, 17711, 6765, 2584, 987, 377, 144, 55, 21

Offset: 1

Emeric Deutsch, Jul 31 2006

Also the number of directed column-convex polyominoes of area n, having k cells in the last column. Row sums are the odd-subscripted Fibonacci numbers (A001519). Sum_{k=1..n} k*T(n,k) = Fibonacci(2n) = A001906(n).
Riordan array ((1-2*x+x^2)/(1-3*x+x^2), x). - Philippe Deléham, Oct 04 2014
Antidiagonal sums are in A007598. - Philippe Deléham, May 22 2015

			T(4,2)=3 because we have UUDD(UU)DD, UUD(UU)DDD and UDUD(UU)DD, where U=(1,1) and D=(1,-1) (the last ascents are shown between parentheses).
Triangle starts:
   1;
   1,  1;
   3,  1, 1;
   8,  3, 1, 1;
  21,  8, 3, 1, 1;
  55, 21, 8, 3, 1, 1;
  ...

E. Barcucci, A. Del Lungo, S. Fezzi and R. Pinzani, Nondecreasing Dyck paths and q-Fibonacci numbers, Discrete Math., 170, 1997, 211-217.
E. Barcucci, R. Pinzani and R. Sprugnoli, Directed column-convex polyominoes by recurrence relations, Lecture Notes in Computer Science, No. 668, Springer, Berlin (1993), pp. 282-298.
A. M. Baxter, L. K. Pudwell, Ascent sequences avoiding pairs of patterns, 2014.
E. Deutsch and H. Prodinger, A bijection between directed column-convex polyominoes and ordered trees of height at most three, Theoretical Comp. Science, 307, 2003, 319-325.

Cf. A001519, A001906, A104762, A088305.

Cf. A000045, A007598.

Maple
```
with(combinat): T:=proc(n,k) if k
				
```

T(n,k) = Fibonacci(2(n-k)) if k < n; T(n,n)=1.
G.f.: G = G(t,z) = t*z*(1-z)^2/((1-3z+z^2)*(1-tz)).
From Gary W. Adamson, Jul 07 2011: (Start)
Let M be the production matrix:
  1, 1, 0, 0, 0, 0, ...
  2, 0, 1, 0, 0, 0, ...
  3, 0, 0, 1, 0, 0, ...
  4, 0, 0, 0, 1, 0, ...
  5, 0, 0, 0, 0, 1, ...
  ...
n-th row of triangle A121461 = top row terms of (n-1)-th power of M. (End)
Let P denote Pascal's triangle. Then P^(-1)*A121461*P = A104762. - Peter Bala, Apr 11 2013

A129170 Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n such that the sum of the height of the peaks is k (n >= 0; n <= k <= floor((n+1)^2/4)).

Original entry on oeis.org

1, 0, 1, 0, 0, 3, 0, 0, 0, 8, 2, 0, 0, 0, 0, 21, 9, 6, 0, 0, 0, 0, 0, 55, 33, 29, 16, 4, 0, 0, 0, 0, 0, 0, 144, 111, 114, 84, 60, 18, 12, 0, 0, 0, 0, 0, 0, 0, 377, 355, 409, 356, 305, 199, 120, 58, 32, 8, 0, 0, 0, 0, 0, 0, 0, 0, 987, 1098, 1389, 1365, 1308, 1032, 843, 507, 372, 204, 120

Offset: 0

Emeric Deutsch, Apr 07 2007

A skew Dyck path is a path in the first quadrant which begins at the origin, ends on the x-axis, consists of steps U=(1,1)(up), D=(1,-1)(down) and L=(-1,-1)(left) so that up and left steps do not overlap. The length of the path is defined to be the number of its steps.
Row n has 1 + floor((n+1)^2/4) terms, the first n of which are equal to 0.
Row sums yield A002212.

			T(3,4)=2 because we have UUDUDD and UUDUDL.
Triangle starts:
  1;
  0,  1;
  0,  0,  3;
  0,  0,  0,  8,  2;
  0,  0,  0,  0, 21,  9,  6;
  0,  0,  0,  0,  0, 55, 33, 29, 16,  4;

E. Deutsch, E. Munarini, S. Rinaldi, Skew Dyck paths, J. Stat. Plann. Infer. 140 (8) (2010) 2191-2203

Cf. A001906, A002212, A088305, A129171.

H:=(1+z*h[1]-z)/(1-z*h[1]+z-z*t*x): for n from 1 to 11 do h[n]:=(1+z*h[n+1]-z)/(1-z*h[n+1]+z-z*t^(n+1)*x) od: h[12]:=0: x:=1: G:=simplify(H): Gser:=simplify(series(G,z=0,11)): for n from 0 to 9 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 9 do seq(coeff(P[n],t,j),j=0..floor((n+1)^2/4)) od; # yields sequence in triangular form

T(n,n) = A088305(n) = A001906(n) = Fibonacci(2n) for n >= 1.
Sum_{k>=0} k*T(n,k) = A129171(n).
G.f.: G(t,z)=H(t,1,z), where H(t,x,z)=1+z[H(t,tx,z)-1+tx]H(t,x,z)+z[H(t,tx,z)-1] (H(t,x,z) is the trivariate g.f. for skew Dyck paths according to sum of the height of the peaks, number of peaks and semilength, marked by t,x and z, respectively).

A204533 Triangle T(n,k), read by rows, given by (0, 1, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 3, 2, 1, 0, 8, 7, 3, 1, 0, 21, 22, 12, 4, 1, 0, 55, 67, 43, 18, 5, 1, 0, 144, 200, 147, 72, 25, 6, 1, 0, 377, 588, 486, 271, 110, 33, 7, 1, 0, 987, 1708, 1566, 976, 450, 158, 42, 8, 1

Offset: 0

Philippe Deléham, Jan 16 2012

Riordan array (1, x*(1-x)^2/(1-3*x+x^2)).

Antidiagonal sums: see A052946.

			Triangle begins:
  1;
  0,   1;
  0,   1,   1;
  0,   3,   2,   1;
  0,   8,   7,   3,  1;
  0,  21,  22,  12,  4,  1;
  0,  55,  67,  43, 18,  5, 1;
  0, 144, 200, 147, 72, 25, 6, 1;

Michael De Vlieger, Table of n, a(n) for n = 0..11475 (rows 0 <= n <= 150).

Cf. diagonals: A000007, A088305, A000012, A001477, A055998.

Table[Sum[Binomial[k, m - 1] Binomial[n - 2 m + k, n - k - 1], {k, 0, n - 1}] + Boole[n == m == 0], {n, 0, 9}, {m, 0, n}] // Flatten (* Michael De Vlieger, Sep 26 2018 *)

T(n,m):= if n=0 and m=0 then 1 else sum(binomial(k,m-1)*binomial(n-2*m+k,n-k-1),k,0,n-1); /* Vladimir Kruchinin, Sep 27 2018 */

T(n,k) = if ((n==0) && (k==0), 1, sum(i=0, n-1, binomial(i,k-1)*binomial(n-2*k+i,n-i-1))); \\ Michel Marcus, Sep 27 2018

Sum_{k=0..n} T(n,k) = A204200(n+1).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) + T(n-3,k-1) - T(n-2,k) - 2*T(n-2,k-1).
G.f.: (-1 + 3*x - x^2)/(-1 + 3*x - x^2 + x*y - 2*x^2*y + x^3*y). - R. J. Mathar, Aug 11 2015
T(n,m) = Sum_{k=0..n-1} C(k,m-1)*C(n-2*m+k,n-k-1), T(0,0)=1. - Vladimir Kruchinin, Sep 27 2018

A208736 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with exactly 2 elements of each rank level between 0 and 1.

Original entry on oeis.org

0, 0, 0, 1, 5, 22, 91, 361, 1392, 5265, 19653, 72694, 267179, 977593, 3565600, 12975457, 47142021, 171075606, 620303547, 2247803785, 8141857808, 29481675889, 106728951109, 386314552438, 1398132674955, 5059626441177, 18308871648576, 66249898660801

Offset: 0

David Nacin, Mar 01 2012

Uniform used in the sense of Retakh, Serconek and Wilson. We use Stanley's definition of graded poset: all maximal chains have the same length n (which also implies all maximal elements have maximal rank.)

R. Stanley, Enumerative combinatorics. Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. Retakh, S. Serconek, and R. Wilson, Hilbert Series of Algebras Associated to Directed Graphs and Order Homology, arXiv:1010.6295 [math.RA], 2010-2011.
Wikipedia, Graded poset
Index entries for linear recurrences with constant coefficients, signature (8,-21,20,-5).

Cf. A208737, A206901, A206902, A206947-A206950, A001906, A025192, A081567, A124302, A124292, A088305, A086405, A012781.

Join[{0, 0}, LinearRecurrence[{8, -21, 20, -5}, {0, 1, 5, 22}, 40]]

def a(n, d={0:0,1:0,2:0,3:1,4:5,5:22}):
    if n in d:
        return d[n]
    d[n]=8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4)
    return d[n]

a(n) = 8*a(n-1) - 21*a(n-2) + 20*a(n-3) - 5*a(n-4), a(2) = 0, a(3) = 1, a(4) = 5, a(5) = 22.
G.f.: (x^3 - 3*x^4 + 3*x^5)/(1 - 8*x + 21*x^2 - 20*x^3 + 5*x^4); (x^3 * (1 - 3*x + 3*x^2))/((1 - 3*x + x^2)*(1 - 5*x + 5*x^2)) .
a(n) = A081567(n-2) - A001519(n-1).

A208737 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with no 3-element antichain.

Original entry on oeis.org

0, 0, 0, 1, 7, 37, 175, 778, 3325, 13837, 56524, 227866, 909832, 3607294, 14227447, 55894252, 218937532, 855650749, 3338323915, 13007422705, 50631143323, 196928737582, 765495534433, 2974251390529, 11552064922624, 44856304154086

Offset: 0

David Nacin, Mar 01 2012

Uniform used in the sense of Retakh, Serconek and Wilson. We use Stanley's definition of graded poset: all maximal chains have the same length n (which also implies all maximal elements have maximal rank.)

R. Stanley, Enumerative combinatorics. Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. Retakh, S. Serconek, and R. Wilson, Hilbert Series of Algebras Associated to Directed Graphs and Order Homology, arXiv:1010.6295 [math.RA], 2010-2011.
Wikipedia, Graded poset
Index entries for linear recurrences with constant coefficients, signature (10,-36,57,-39,9).

Cf. A208736, A206901, A206902, A206947-A206950, A001906, A025192, A081567, A124302, A124292, A088305, A086405, A012781.

Join[{0}, LinearRecurrence[{10, -36, 57, -39, 9}, {0, 0, 1, 7, 37}, 40]]

def a(n, d={0:0,1:0,2:0,3:1,4:7,5:37}):
    if n in d:
        return d[n]
    d[n]=10*a(n-1) - 36*a(n-2) + 57*a(n-3) - 39*a(n-4) + 9*a(n-5)
    return d[n]

a(n) = 10*a(n-1) - 36*a(n-2) + 57*a(n-3) - 39*a(n-4) + 9*a(n-5), a(1) = 0, a(2) = 0, a(3) = 1, a(4) = 7, a(5) = 37.
G.f: (x^3 - 3*x^4 + 3*x^5)/(1 - 10*x + 36*x^2 - 57*x^3 + 39*x^4 - 9*x^5); (x^3*(1 - 3*x + 3*x^2)) / ((1 - x) (1 - 3*x) (1 - 6*x + 9*x^2 - 3*x^3)).
a(n) = A124292(n) - A124302(n).

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

A307415 Total number of parts in all symmetric m-color cyclic compositions of n (that is, the total number of parts in all achiral cyclic compositions of n where a part with size m can be colored with one of m colors).

Original entry on oeis.org

1, 4, 10, 26, 53, 116, 215, 434, 766, 1480, 2539, 4776, 8045, 14864, 24722, 45094, 74305, 134236, 219619, 393790, 640646, 1141844, 1849175, 3279696, 5291353, 9346396, 15031450, 26458994, 42438221, 74479940, 119182319, 208629386, 333170830, 581904544, 927617347, 1616924664

Offset: 1

Petros Hadjicostas, Jun 24 2019

nonn

Cyclic compositions of a positive integer n are equivalence classes of ordered partitions of n such that two such partitions are equivalent if one can be obtained from the other by rotation. These were first studied by Sommerville (1909).
Symmetric cyclic compositions or circular palindromes or achiral cyclic compositions are those cyclic compositions that have at least one axis of symmetry. They were also studied by Sommerville (1909, pp. 301-304).
Let c = (c(m): m >= 1) be the input sequence and let b_k = (b_k(n): n >= 1) be the output sequence under Bower's CPAL[k] (circular palindrome with k boxes) transform of c; that is, b_k(n) = (CPAC[k] c)n for n >= 1. Hence, b_k(n) is the number of symmetric cyclic compositions of n with k parts, where a part of size m can be colored with one of c(m) colors. If C(x) = Sum{m >= 1} c(m)*x^m is the g.f. of the input sequence c, then the bivariate g.f. of the list of sequences (b_k: k >= 1) = ((CPAL[k] c): k >= 1) = ((b_k(n): n >= 1): k >= 1) is Sum_{n,k >= 1} b_k(n)*x^n*y^k = (1 + y*C(x))^2/(2 * (1 - y^2*C(x^2))) - (1/2).
Here, Sum_{k=1..n} k*b_k(n) is the total number of parts in all symmetric colored cyclic compositions of n (where the coloring of parts is done according to the input sequence c). To find the g.f. of the sequence (Sum_{k=1..n} k*b_k(n): n >= 1), we differentiate the above bivariate g.f. (i.e., Sum_{n,k >= 1} b_k(n)*x^n*y^k) w.r.t. y and set y = 1. We get (1 + C(x)) * (C(x) + C(x^2))/(1 - C(x^2))^2.
For the current sequence, the input sequence is c(m) = m for m >= 1, and we are dealing with the so-called "m-color" compositions. m-color linear compositions were studied by Agarwal (2000), whereas m-color cyclic compositions were studied by Gibson (2017) and Gibson et al. (2018).
Thus, for the current sequence, a(n) is the total number of parts in all symmetric (achiral) m-color cyclic compositions of n (where a part of size m may be colored with one of m colors). Since C(x) = x/(1 - x)^2, we get that (1 + C(x)) * (C(x) + C(x^2))/(1 - C(x^2))^2 = (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2).
The roots of the denominator (x^2 + x - 1)^2 * (x^2 - x - 1)^2 are phi, -phi, 1/phi, -1/phi, where phi = (1 + sqrt(5))/2 = A001622 is the golden ratio. Each of these roots is double, so a(n) = (c_1 + d_1*n) * phi^n + (c_2 + d_2*n) * (-phi)^n + (c_3 + d_3*n) * (1/phi)^n + (c_4 + d_4*n) * (-1/phi)^n, and the constants c_i, d_i (i = 1, 2, 3, 4) can be determined from the first eight terms of (a(n): n >= 1). Since, however, the Fibonacci numbers (A000045(n): n >= 0) can be expressed in terms of phi, after some tedious algebra, we may derive the formula given below.

			We have a(1) = 1 because we only have one symmetric cyclic composition of n = 1, namely 1_1, which has 1 part.
We have a(2) = 4 because we have the following colored achiral cyclic compositions of n = 2: 2_1, 2_2, 1_1 + 1_1; hence, a(2) = 1 + 1 + 2 = 4.
We have a(3) = 10 because we have the following colored achiral cyclic compositions of n = 3: 3_1, 3_2, 3_3, 1_1 + 2_1, 1_1 + 2_2, 1_1 + 1_1 + 1_1; hence, a(3) = 1 + 1 + 1 + 2 + 2 + 3 = 10.
We have a(4) = 26 because we have the following colored achiral cyclic compositions of n = 4: 4_1, 4_2, 4_3, 4_4, 1_1 + 3_1, 1_1 + 3_2, 1_1 + 3_3, 2_1 + 2_1, 2_1 + 2_2, 2_2 + 2_2, 1_1 + 2_1 + 1_1, 1_1 + 2_2 + 1_1, 1_1 + 1_1 + 1_1 + 1_1; hence, a(4) = 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + 4 = 26.
We have a(5) = 53 because we have the following colored achiral cyclic compositions of n = 5:
(i) With one part: 5_1, 5_2, 5_3, 5_4, 5_5; i.e., a total of 5 parts.
(ii) With two parts: 1_1 + 4_1, 1_1 + 4_2, 1_1 + 4_3, 1_1 + 4_4, 2_1 + 3_1, 2_1 + 3_2, 2_1 + 3_3, 2_2 + 3_1, 2_2 + 3_2, 2_2 + 3_3; i.e., a total of 20 parts.
(iii) With three parts: 1_1 + 3_1 + 1_1, 1_1 + 3_2 + 1_1, 1_1 + 3_3 + 1_1, 2_1 + 1_1 + 2_1, 2_2 + 1_1 + 2_2; i.e., a total of 15 parts.
(iv) With four parts: 1_1 + 1_1 + 2_1 + 1_1, 1_1 + 1_1 + 2_2 + 1_1 (here, the axis of symmetry passes through one of the 1's and through 2); i.e., a total of 8 parts.
(v) With five parts: 1_1 + 1_1 + 1_1 + 1_1 + 1_1; i.e., a total of 5 parts.
Thus, a(5) = 5 + 20 + 15 + 8 + 5 = 53.

A. K. Agarwal, n-colour compositions, Indian J. Pure Appl. Math. 31(11) (2000), 1421-1427.
C. G. Bower, Transforms (2).
Petros Hadjicostas, Generalized colored circular palindromic compositions, Moscow Journal of Combinatorics and Number Theory, 9(2) (2020), 173-186.
Meghann Moriah Gibson, Combinatorics of compositions, Master of Science, Georgia Southern University, 2017.
Meghann Moriah Gibson, Daniel Gray, and Hua Wang, Combinatorics of n-color compositions, Discrete Mathematics 341 (2018), 3209-3226.
D. M. Y. Sommerville, On certain periodic properties of cyclic compositions of numbers, Proc. London Math. Soc. S2-7(1) (1909), 263-313.

Cf. A000045, A001622, A030267, A032198, A032287, A088305, A308723.

a(n) = (n/5) * (11*F(n) + 7*F(n-1)) + (-1)^n * (n/5) * (-4*F(n) + 7*F(n-1)) - (2/5) * (5*F(n+1) + 3*F(n)) - (-1)^n * (2/5) * (3*F(n) - 5*F(n-1)) for n >= 1, where F(n) = A000045(n) is the n-th Fibonacci number.

G.f.: (x^2 - x + 1) * x * (x^2 + 3*x + 1) * (x + 1)^2/((x^2 + x - 1)^2 * (x^2 - x - 1)^2).

A353164 Expansion of 1/(1 - Sum_{p prime} p * x^p).

Original entry on oeis.org

1, 0, 2, 3, 4, 17, 17, 63, 100, 211, 495, 846, 2057, 3831, 8181, 17078, 33788, 72705, 144801, 303452, 623115, 1274365, 2652052, 5408046, 11207927, 23020231, 47378495, 97774736, 200819019, 414365805, 852285510, 1755453858, 3616014678, 7441523271, 15332278869

Offset: 0

Seiichi Manyama, Apr 28 2022

nonn

Cf. A000040, A023360, A088305, A353162, A353165.

my(N=40, x='x+O('x^N)); Vec(1/(1-sum(k=1, N, isprime(k)*k*x^k)))

a(n) = if(n==0, 1, sum(k=1, n, isprime(k)*k*a(n-k)));

a(0) = 1; a(n) = Sum_{p<=n, p prime} p * a(n-p).

cp's OEIS Frontend

A329157 Expansion of Product_{k>=1} (1 - Sum_{j>=1} j * x^(k*j)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A382992 Number of compositions of n that have at least 1 part equal to 1 and any part 1 at position k can be k different colors.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

Extensions

A121461 Triangle read by rows: T(n,k) is the number of nondecreasing Dyck paths of semilength n, having last ascent of length k (1 <= k <= n).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

A129170 Triangle read by rows: T(n,k) is the number of skew Dyck paths of semilength n such that the sum of the height of the peaks is k (n >= 0; n <= k <= floor((n+1)^2/4)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Formula

A204533 Triangle T(n,k), read by rows, given by (0, 1, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Maxima

PARI

Formula

A208736 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with exactly 2 elements of each rank level between 0 and 1.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Mathematica

Python

Formula

A208737 Number of nonisomorphic graded posets with 0 and 1 and non-uniform Hasse graph of rank n, with no 3-element antichain.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs